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Q1. Sketch the graph y = |x + 1|. Evaluate |x + 1|dx. What does the value of this integral represent on this graph? Ans1.
y = |x + 1| = {(x + 1) if x > - 1} and {- (x + 1) if x < -1}Graph is a pair of lines through x = -1 making angle 45o & 135o with x-axis because slope of the lines are + 1. Further to draw the lines we need two points on each lines. Join the pair to point (-1, 0) and (0, 1) by line to get graph to y = x + 1. Also join the pair of points (-1, 0) and (-2, 1) by line to get graph of y = -(x + 1).Evaluation of integral.
= -[-1/2 - 4] + [4 + 1/2] = 1/2 + 4 + 4 + 1/2 = 9
Representation of the value 9 of integral on graph. |x + 1|dx = 9 represent the area bounded by the curve y = |x + 1|, X-axis and the lines x = -4 and x = 2 i.e. it is equal to the sum of the area of triangle ABD and ACE. = (1/2)(9) + (1/2)(9) = 9/2 + 9/2 = 9 [... Area = 1/2 base x height] Q2. Find the area bounded by x = at2, y = 2 at between ordinates corresponding to t = 1 and t = 2. Ans2. Eliminate 't' form give two relationy = 2at t = y/2a x = at2 = a(y/2a)2
= y2/4a y2 = 4ax ..............(1)Also t = 1 x = a and y = 2at = 2 x = 4a and y = 4aNow (1) is parabola which is symmetrical about x-axis
Q3. Compute the area bounded by the lines x + 2y = 2, y - x = 1 & 2x + y = 7. Ans3.
Given lines are x + 2y.......(1), y - x = 1 ...........(2), 2x + y = 7 ........(3)Line (1), meets x-axis at (2, 0) and y-axis at (0, 1) Join (2, 0) and (0, 1) to get the graph of first lines. Similarly second lines is join of (-1, 0) and (0, 1) and third line is join (0, 7) and (7/2, 0)Lines (1) and (2) meet in (0, 1)Lines (2) and (3) meet in (2, 3)Lines (3) and (1) meet in (4, -1).Thus the points of intersection of the three lines are A(0,1), B(2,3), C(4, -1). Area bounded by the lines is shown shaded in the figure.Required area = Area ABC= Area ABD + Area DEB + Area DLE + Area LCE} .........(4)Now Area ADB = Area DBAO - Area DAO
= Modulus (-3/16) = 3/16......(8)Using the values of areas obtained (5), (6), (7), (8) in (4) we haveRequired area = 3 + 9/4 + 9/16 + 3/16 = 96/16 = 6 sq. units Q4. Sketch the graph y = (x) + 1 in {0, 4} and determine the area of the region enclosed by the curve, the axis of x and the lines x = 0, x = 4. Ans4.
y = (x) + 1 x = y - 1 x = (y - 1)2.........(1)Shift origin to the point (0, 1) x = X + 0 and y = Y + 1 put (1) we getX = Y2 which is parabola (Right Handed) with vertex at (0,1).Plot these points and join them by free hand curve to get rough sketchRequired Area (Shaded)
= 2/3 (8) + 4 = 16/3 + 4
= 28/3 sq.units Q5. Find the area of the region{(x, y) : y2 4x, 4x2 + 4y2 9} Ans5.
Let R = {(x, y):y2 4x, 4x2 + 4y2 9}= {x, y:y2 4x} {(x, y):4x2 + 4y2 < 9}= R1 R2Where R1 = {x, y:y2 4x} represent the region inside the parabola y2 = 4x with vertex (0, 0) and x-axis as it axis.and R2 = {x , y:4x2 + 4y2 < 9} represent the interior of the circle 4x2 + 4y2 = 9 with centre (0, 0) and radius (3/2).Thus the region R which is intersection of R1 & R2 is shown shaded in the figure.Now to find the points intersection of the given curves, we solve their equation y2 = 4x ...........(1) and 4x2 + 4y2 = 9 ..........(2) simultaneously.Using (1) & (2) we get4x2 + 4(4x) = 9 4x2 + 16x - 9 = 04x2 + 18x - 2x - 9 = 0(2x - 1) (2x + 9) = 0x = 1/2, -9/2From (1) x = 1/2 y = + 2 and x = -9/2 y = imaginary quantityThus the points of intersection of (1) and (2) are A(1/2, 2) and B (1/2, 2)Also both the curve are symmetrical about x-axis Required Area = Area OBPAO - 2.Area (OPAO)= 2[Area ODA + Area DPA]
= 8/3(1/22) + (9/4)/2 - (1/2) - 9/4 sin-1 (1/3)= (22/3) - 1/2 + 9/8 - 9/4 sin-1 (1/3) sq.units Q6. Find the area of the region enclosed between the two circle x2 + y2 = 1 and (x - 1)2 + y2 = 1 Ans6.
x2 + y2 = 1............(1)(x - 1)+ y2 = 1.........(2)Intersect at the point obtained by solving (1) and (2)From (1) y2 = 1 - x2, putting in (2) we get(x - 1)2 + 1 - x2 = 1 -2x + 1 = 0 x = 1/2From (1) x = 1/2 y = + 3/2 (1) and (2) intersect at A[1/2, 3/2] and B [1/2, 3/2] Center of first circle is (0, 0) and radius = 1Also centre of second circle is (1, 0) and radius = 1Also both, the circle are symmetrical about x-axis Required area is shown shaded Required area = Area OACB= 2(Area OAC) = 2[Area OAD + Area DCA]
= (2/3 - 3/2) sq. units Q7. Find the area of the region {(x, y)}:x2 + y2 1 x + y}. Ans7.
The required area is the area between the circle x2 + y2 = 1 and line x + y = 1. Circle (1) has centre (0, 0) and radius and line to meets x-axis at A (1, 0) y-axis at B (0, 1). The circle (1) also passes through A and B. Hence points of intersection of (1) and (2) are A (1, 0) and B (0, 1)
= (1/2) sin-1(1) - 1/2= (1/2)(/2) - 1/2= (/4 - 1/2) sq.units Q8. Using integration find the area of ABC with vertices at A (2, 5) B (4, 7) and C (6, 2). Ans8.
Equation the AB of ABC is(y - 5) = [{(7 - 5)/(4 - 2)}(x - 2)]y = x + 3Similarly equation of sides BC and CA are 5x + 2y - 34 = 0 and 3x + 4y - 26 = 0Required Area = Area of ABC= Area LMBA + Area of MNCB - Area LNCA